I've recently begun reading up on differential forms in a physics context, and my resources said that one can often derive vector identities from differential forms.
For instance,
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$\nabla \cdot (\nabla \times F)=0$ and $\nabla \times (\nabla F)=0$ both follow from $\text{dd}=0$, where $\text{d}$ is the exterior derivative.
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$A\cdot (B\times C)=B\cdot (C\times A)=C\cdot (A\times B)$ follows from $A \wedge B \wedge C=B \wedge C \wedge A=C \wedge A \wedge B$, where $\wedge$ is the wedge product.
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$\nabla \cdot (A\times B)=B\cdot (\nabla\times A)-A\cdot(\nabla\times B)$ follows from the product rule for the exterior derivative: $d(A\wedge B)=dA\wedge B + (-1)^P A\wedge dB$ (where $P$ is the degree of $A$).
I tried deriving the infamous BAC-CAB rule,
$A\times(B\times C)= B(A\cdot C) – C(A\cdot B)$, in this mode, but I couldn't get too far.
I would think the relevant expression in the language of differential forms should perhaps be $*(A\wedge *(B\wedge C)) = (B\wedge (A\lrcorner C) – C\wedge (A\lrcorner B))$, but I'm not sure. (Where $*$ is the Hodge star operator and $\lrcorner$ is the interior product).
Does anyone have some insight to share on deriving this identity or others from differential forms? Thanks!
Best Answer
I know that this question already has an accepted answer, but let me give an answer from a purist differential forms perspective. Recall that if $V$ is an $n$-dimensional inner product space, then the Hodge star is a linear isomorphism $\ast : \bigwedge^k V \to \bigwedge^{n-k} V$ for each $0 \leq k \leq n$, satisfying the following:
for $v$, $w \in V$, $\langle v,w\rangle \omega = v \wedge \ast w$ for $\omega = \ast 1$ the generator of $\bigwedge^n V$ satisfying $\omega = e_1 \wedge \cdots \wedge e_n$ for any orthonormal basis $\{e_k\}$ of $V$ with the appropriate orientation (e.g., the volume form in $\bigwedge^n (\mathbb{R}^n)^\ast$);
in particular, $\ast \ast = \operatorname{Id}$ when $n$ is odd.
Also, recall that the inner product on $\bigwedge^k V$ is given by $$ \langle v_1 \wedge \cdots \wedge v_k, w_1 \wedge \cdots \wedge w_k \rangle = \det(\langle v_i,w_j \rangle). $$
So, suppose that $V$ is $3$-dimensional, in which case we can define the cross product of $a$, $b \in V$ by $$ a \times b := \ast (a \wedge b). $$ Let $u$, $v$, $w \in V$. Then for any $x \in V$, $$ \langle u \times (v \times w), x \rangle \omega = \langle x, u \times (v \times w) \rangle \omega\\ = \langle x, \ast (u \wedge \ast(v \wedge w)) \rangle \omega\\ = x \wedge \ast \ast (u \wedge \ast(v \wedge w))\\ = x \wedge (u \wedge \ast(v \wedge w))\\ = (x \wedge u) \wedge \ast (v \wedge w)\\ = \langle x \wedge u, v \wedge w \rangle \omega\\ = (\langle x,v \rangle\langle u,w \rangle - \langle x,w \rangle\langle u,v \rangle)\omega\\ = \langle \langle u,w \rangle v - \langle u,v \rangle w, x \rangle \omega, $$ implying that $$ \langle u \times (v \times w), x \rangle = \langle \langle u,w \rangle v - \langle u,v \rangle w, x \rangle, $$ and hence, since $x$ was arbitrary, that $$ u \times (v \times w) = \langle u,w \rangle v - \langle u,v \rangle w, $$ as required.